Parametric Equations

Quick answer

Parametric equations give the coordinates of a point as functions of a third variable, the parameter: x = f(t) and y = g(t). As t runs through its values, the point traces a curve, and the direction it travels is the curve's orientation. Parametric form describes curves that are not graphs of functions, such as circles, and records how a point moves, not only where it goes. Eliminating the parameter recovers an equation in x and y, sometimes with a restricted domain.

What you'll learn

Plot a parametric curve from a table of values, with its direction
Eliminate the parameter to find an equation in x and y
Parametrize circles, ellipses and line segments
Model projectile motion with parametric equations

A point that moves

An equation like $y = x^2$ says where a curve is. Often we also want to know how a point travels along it. Parametric equations give both coordinates as functions of a third variable $t$ , the parameter:

x = f(t) \qquad\qquad y = g(t)

Think of $t$ as time. At each moment the point is at $\big(f(t), g(t)\big)$ , and as $t$ increases, the point traces a curve.

To plot one, make a table. For $x = t - 1$ and $y = t^2$ :

$t$	$-2$	$-1$	$0$	$1$	$2$
$x$	$-3$	$-2$	$-1$	$0$	$1$
$y$	$4$	$1$	$0$	$1$	$4$

x = t − 1, y = t²

The path of x = t − 1, y = t² for −2 ≤ t ≤ 2

The direction of travel is the curve’s orientation, marked with an arrow.

Why a parameter says more than an equation

Three pairs of equations trace the same unit circle:

$x = \cos t$ , $y = \sin t$ goes around once, counterclockwise, as $t$ runs from $0$ to $2\pi$ .
$x = \cos 2t$ , $y = \sin 2t$ goes around twice in the same time: twice as fast.
$x = \cos t$ , $y = -\sin t$ goes around clockwise.

All three satisfy $x^2 + y^2 = 1$ , and that one equation cannot tell them apart. An equation in $x$ and $y$ says where the curve is; parametric equations also say when the point gets there and which way it is going. They also describe curves that fail the vertical line test, such as circles, without splitting them into pieces.

Circles, ellipses and segments

For sines and cosines, eliminate the parameter with $\cos^2 t + \sin^2 t = 1$ instead of solving for $t$ . A few parametrizations come up constantly:

Curve	Parametric equations
circle, center $(h, k)$ , radius $r$	$x = h + r\cos t$ , $y = k + r\sin t$
ellipse with semi-axes $a$ and $b$	$x = a\cos t$ , $y = b\sin t$
segment from $(x_1, y_1)$ to $(x_2, y_2)$	$x = x_1 + (x_2 - x_1)t$ , $y = y_1 + (y_2 - y_1)t$ , $0 \le t \le 1$

Worked examples

Common mistakes

Practice problems

For $x = 2t$ , $y = t + 1$ , find the points at $t = 0$ , $1$ and $2$ , and eliminate the parameter.

Answer

$(0, 1)$ , $(2, 2)$ , $(4, 3)$ ; $y = \tfrac{x}{2} + 1$

Full solution

$t = \tfrac{x}{2}$ , so $y = \tfrac{x}{2} + 1$ , a line.
Eliminate the parameter: $x = t + 2$ , $y = t^2$ .

Answer

$y = (x - 2)^2$

Full solution

$t = x - 2$ , and substituting gives $y = (x - 2)^2$ .
Eliminate the parameter: $x = 4\cos t$ , $y = 4\sin t$ .

Answer

$x^2 + y^2 = 16$

Full solution

$x^2 + y^2 = 16\cos^2 t + 16\sin^2 t = 16$ , a circle of radius $4$ .
Eliminate the parameter: $x = 2\cos t$ , $y = 5\sin t$ .

Answer

$\tfrac{x^2}{4} + \tfrac{y^2}{25} = 1$

Full solution

$\left(\tfrac{x}{2}\right)^2 + \left(\tfrac{y}{5}\right)^2 = \cos^2 t + \sin^2 t = 1$ , an ellipse.
Eliminate the parameter: $x = \sqrt{t}$ , $y = 2t$ .

Answer

$y = 2x^2$ for $x \ge 0$

Full solution

$t = x^2$ , so $y = 2x^2$ . Since $x = \sqrt{t} \ge 0$ , only the right half appears.
Which way does $x = \sin t$ , $y = \cos t$ travel around the unit circle as $t$ increases?

Answer

Clockwise

Full solution

At $t = 0$ the point is at $(0, 1)$ , and at $t = \tfrac{\pi}{2}$ it is at $(1, 0)$ : from the top to the right, which is clockwise.
Parametrize the segment from $(-1, 3)$ to $(4, 0)$ .

Answer

$x = -1 + 5t$ , $y = 3 - 3t$ , $0 \le t \le 1$

Full solution

The change is $(5, -3)$ . Start at $(-1, 3)$ and add $t$ times the change.
A ball is kicked from the ground with $x = 30t$ and $y = 20t - 4.9t^2$ , in meters and seconds. How long is it in the air, and how far does it travel?

Answer

About $4.08$ seconds and $122$ meters

Full solution

$y = t(20 - 4.9t) = 0$ at $t = 0$ and $t = \tfrac{20}{4.9} \approx 4.08$ . Then $x = 30 \cdot 4.08 \approx 122.4$ .
Where does the curve $x = t^2 - 4$ , $y = t^3 - t$ cross the $y$ -axis?

Answer

At $(0, 6)$ and $(0, -6)$

Full solution

$x = 0$ when $t = \pm 2$ . At $t = 2$ , $y = 8 - 2 = 6$ ; at $t = -2$ , $y = -8 + 2 = -6$ .
A student eliminates the parameter from $x = t^2$ , $y = t^4$ and says the curve is the whole parabola $y = x^2$ . What went wrong?

Hint

Can $x = t^2$ be negative?

Answer

Since $x = t^2 \ge 0$ , the curve is only the right half, $y = x^2$ for $x \ge 0$ .

Full solution

The substitution $y = (t^2)^2 = x^2$ is right, but it forgets which $x$ -values occur. Every $t$ gives $x \ge 0$ , so the left half of the parabola is never reached.

As $t$ runs from $-1$ to $1$ , the point comes down the right half to the origin and goes back up the same way.

Frequently asked questions

What are parametric equations?

A pair of equations x = f(t) and y = g(t) that give both coordinates of a point in terms of a third variable, the parameter t.

How do I eliminate the parameter?

Solve one equation for t and substitute into the other. For sines and cosines, use sin²t + cos²t = 1 instead.

What is the orientation of a parametric curve?

The direction the point moves as t increases. It is often marked with an arrow on the curve.

How do I parametrize the segment from (x₁, y₁) to (x₂, y₂)?

x = x₁ + (x₂ − x₁)t and y = y₁ + (y₂ − y₁)t for 0 ≤ t ≤ 1.

Why use parametric equations at all?

They describe curves that fail the vertical line test, and they record motion: where a point is at each time, and in which direction it travels.

What to learn next

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